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Photoelectrons are emitted with maximum velocity $\mathrm{v}$ when light of frequency $3 \mathrm{f}$ incidents on a photosensitive material of work function $2 \mathrm{hf}$. If the frequency of the incident light is $4.25 \mathrm{f}$, the maximum velocity of the emitted photoelectrons is
(h-Plank's constant)
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(h-Plank's constant)
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Verified Answer
The correct answer is:
$1.5 \mathrm{v}$
Initial frequency, $f_1=3 f$
Initial velocity $\mathrm{v}_1=\mathrm{v}$
Work function $\phi=2 \mathrm{hf}$
final frequency, $\mathrm{f}_2=4.25 \mathrm{f}$
final velocity, $\mathrm{v}_2=$ ?
Kinetic energy is given as,
$K E=h f-\phi$
$\frac{1}{2} \mathrm{mv}_1^2=3 \mathrm{hf}-2 \mathrm{hf}=\mathrm{hf}$
$m=\frac{2 h f}{v^2}$
final kinetic energy,
$\begin{aligned} & \frac{1}{2} \mathrm{mv}^2=4.25 \mathrm{hf}-2 \mathrm{hf} \\ & =2.25 \mathrm{hf}\end{aligned}$
$\begin{aligned} & \frac{1}{2} \times \frac{2 h f}{v^2} \times v_2^2=\frac{225}{100} \\ & \frac{v_2^2}{v^2}=\frac{225}{100} \\ & v_2^2=\frac{225}{100} v^2=\left(\frac{15}{10} v\right)^2 \\ & v_2=1.5 v\end{aligned}$
Initial velocity $\mathrm{v}_1=\mathrm{v}$
Work function $\phi=2 \mathrm{hf}$
final frequency, $\mathrm{f}_2=4.25 \mathrm{f}$
final velocity, $\mathrm{v}_2=$ ?
Kinetic energy is given as,
$K E=h f-\phi$
$\frac{1}{2} \mathrm{mv}_1^2=3 \mathrm{hf}-2 \mathrm{hf}=\mathrm{hf}$
$m=\frac{2 h f}{v^2}$
final kinetic energy,
$\begin{aligned} & \frac{1}{2} \mathrm{mv}^2=4.25 \mathrm{hf}-2 \mathrm{hf} \\ & =2.25 \mathrm{hf}\end{aligned}$
$\begin{aligned} & \frac{1}{2} \times \frac{2 h f}{v^2} \times v_2^2=\frac{225}{100} \\ & \frac{v_2^2}{v^2}=\frac{225}{100} \\ & v_2^2=\frac{225}{100} v^2=\left(\frac{15}{10} v\right)^2 \\ & v_2=1.5 v\end{aligned}$
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